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Question Number 155294 by peter frank last updated on 28/Sep/21

Answered by peter frank last updated on 30/Sep/21

Commented by peter frank last updated on 30/Sep/21

$$\mathrm{Point}\mathrm{P}=\mathrm{point}\mathrm{B}$$$${\mathrm{R}}_1=\frac{{\mathrm{v}}_{\mathrm{o}}^2\sin 2{\theta }_1}{\mathrm{g}}=0.899\mathrm{m}{\theta }_1=46$$$${\mathrm{R}}_2=\frac{{\mathrm{v}}_{\mathrm{o}}^2\sin 2{\theta }_2}{\mathrm{g}}=0.899\mathrm{m}{\theta }_2=44$$$${\mathrm{R}}_1={\mathrm{R}}_2\mathrm{hencd}\mathrm{both}\mathrm{particle}$$$$\mathrm{focused}\mathrm{on}\mathrm{the}\mathrm{same}\mathrm{point}\mathrm{P}$$$$\left(\mathrm{b}\right)\mathrm{h}={\mathrm{h}}_2-{\mathrm{h}}_1$$$${\mathrm{h}}_2=\frac{{\mathrm{v}}_{\mathrm{o}}^2\sin {}^2{\theta }_2}{2\mathrm{g}}$$$${\mathrm{h}}_1=\frac{{\mathrm{v}}_{\mathrm{o}}^2\sin {}^2{\theta }_1}{2\mathrm{g}}$$$$\mathrm{g}=\mathrm{a}=4\times {10}^{13}\mathrm{m}/{\mathrm{s}}^2$$$$\mathrm{hence}$$$$\mathrm{h}=0.233\mathrm{m}$$

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